4: Clustering and classification

date()

## [1] "Mon Nov 27 10:41:04 2023"

Prep packages:

library(dplyr)

## 
## Attaching package: 'dplyr'

## The following objects are masked from 'package:stats':
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##     filter, lag

## The following objects are masked from 'package:base':
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##     intersect, setdiff, setequal, union

library(MASS)

## Warning: package 'MASS' was built under R version 4.3.2

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## Attaching package: 'MASS'

## The following object is masked from 'package:dplyr':
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##     select

library(corrplot)

## Warning: package 'corrplot' was built under R version 4.3.2

## corrplot 0.92 loaded

library(plotly)

## Warning: package 'plotly' was built under R version 4.3.2

## Loading required package: ggplot2

## 
## Attaching package: 'plotly'

## The following object is masked from 'package:ggplot2':
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##     last_plot

## The following object is masked from 'package:MASS':
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##     select

## The following object is masked from 'package:stats':
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## The following object is masked from 'package:graphics':
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##     layout

library(GGally)

## Warning: package 'GGally' was built under R version 4.3.2

## Registered S3 method overwritten by 'GGally':
##   method from   
##   +.gg   ggplot2

Load the Boston data from MASS the R package, explore the structure and the dimensions of the data:

The dataset contains “Housing Values in Suburbs of Boston”.

More information on the data can be found here:

[https://stat.ethz.ch/R-manual/R-devel/library/MASS/html/Boston.html]

data("Boston")

A graphical overview of the data and summaries of the variables in the data:

str(Boston)

## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...

summary(Boston)

##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08205   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

pairs(Boston)

cor_matrix <- cor(Boston) %>% round(2)
corrplot(cor_matrix, method="circle", type = "upper", cl.pos = "b", tl.pos = "d", tl.cex = 0.6)

# ggpairs(Boston, mapping = aes(alpha = 0.3), lower = list(combo = wrap("facethist", bins = 20)))

The distribution of the variabes are different. Some, like zn and age, seem to be from zero to a hundred, some (like chas) from zero to one, and some have complitely different scales: tax from 187 to 711, black from 0.32 to 397.

Many of the variables are heavily correlated. For example, lstat and medv have a negative correlation of almost 1, and rad and tax a positive correlation of almost one.

Standardizing the dataset:

# scale the variables (all numeric)
boston_scaled <- scale(Boston)
summary(boston_scaled)

##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv        
##  Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 3.5453   Max.   : 2.9865

Standardizing helps to unify variables in very different scales. The variables have been scaled and centered. In the standardized data, the mean of each column is zero now.

Create a categorical variable of the crime rate in the Boston dataset:

boston_scaled <- as.data.frame(boston_scaled) # we need this too
boston_scaled$crim <- as.numeric(boston_scaled$crim) # we need this too

crime <- cut(boston_scaled$crim, breaks = quantile(boston_scaled$crim), include.lowest = TRUE, label = c("low", "med_low", "med_high", "high"))

Drop the old crime rate variable from the dataset:

# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)

# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)

Divide the dataset to train and test sets:

# number of rows in the Boston dataset 
n <- nrow(boston_scaled)

# choose randomly 80% of the rows
ind <- sample(n,  size = n * 0.8)

# create train set
train <- boston_scaled[ind,]

# create test set 
test <- boston_scaled[-ind,]

Fitting the linear discriminant analysis on the train set & drawing the LDA (bi)plot:

# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)

# print the lda.fit object
lda.fit

## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2574257 0.2524752 0.2326733 0.2574257 
## 
## Group means:
##                   zn      indus        chas        nox          rm        age
## low       0.96645965 -0.9212144 -0.12090214 -0.8535598  0.44810518 -0.8568763
## med_low  -0.05384482 -0.2874880  0.03646311 -0.5963858 -0.09670717 -0.3880420
## med_high -0.38506488  0.1384192  0.27216352  0.3564530  0.16717690  0.4042563
## high     -0.48724019  1.0170690 -0.08304540  1.0352191 -0.46146119  0.8056106
##                 dis        rad        tax     ptratio       black        lstat
## low       0.8376777 -0.6781900 -0.7505003 -0.46001999  0.38227261 -0.764105556
## med_low   0.4389733 -0.5506331 -0.5179629 -0.06867187  0.35518382 -0.145832181
## med_high -0.3478582 -0.3722063 -0.2875995 -0.31213089  0.04010864  0.001686464
## high     -0.8453773  1.6386213  1.5144083  0.78135074 -0.80175115  0.886882560
##                  medv
## low       0.544682470
## med_low   0.007625754
## med_high  0.203482420
## high     -0.687937968
## 
## Coefficients of linear discriminants:
##                 LD1         LD2         LD3
## zn       0.10648667  0.78594870 -0.79158541
## indus   -0.02441030 -0.28650157  0.64263523
## chas    -0.08397676 -0.09366272  0.12040410
## nox      0.38471916 -0.60328641 -1.32403626
## rm      -0.09623505 -0.12179437 -0.05186723
## age      0.26440199 -0.34572572 -0.20983866
## dis     -0.13520685 -0.39589664  0.37986046
## rad      2.90239847  1.10019954  0.13212773
## tax      0.09649070 -0.25594090  0.24211036
## ptratio  0.10980039  0.02099939 -0.17580804
## black   -0.14564100  0.03061035  0.17310861
## lstat    0.19284745 -0.26240837  0.44047577
## medv     0.17423744 -0.38744108 -0.21591785
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9484 0.0366 0.0150

# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  graphics::arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# target classes as numeric
classes <- as.numeric(train$crime)

# plot the lda results (select both lines and execute them at the same time!)
plot(lda.fit, dimen = 2)
lda.arrows(lda.fit, myscale = 1)

Saving the crime categories from the test set and then removing the categorical crime variable from the test dataset:

# save the correct classes from test data
correct_classes <- test$crime

# remove the crime variable from test data
test <- dplyr::select(test, -crime)

Predicting the classes with the LDA model on the test data:

# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)

# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)

##           predicted
## correct    low med_low med_high high
##   low       15       8        0    0
##   med_low    3      12        9    0
##   med_high   1      11       20    0
##   high       0       0        0   23

Many have been predicted correct (13+16+15+26) / (13+11+1+4+16+3+2+11+15+26) = 0.6862745, in total 68.6% are correct. Although may are still predicted wrong, especially in the low and med_high categories. The model works the best for the high category, where all are correct.

Reloading the Boston dataset and standardizing the dataset:

# from above
data("Boston")
boston_scaled <- scale(Boston)
boston_scaled <- as.data.frame(boston_scaled) # we need this too

Calculating the distances between the observations:

# with euclidean distance

# euclidean distance matrix
dist_eu <- dist(boston_scaled)

# look at the summary of the distances
summary(dist_eu)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.1343  3.4625  4.8241  4.9111  6.1863 14.3970

Running k-means algorithm:

# k-means clustering
km <- kmeans(boston_scaled, centers = 3) # trying with 3 clusters to begin with

# plot the Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)

Investigating the optimal number of clusters and running the algorithm again:

set.seed(123)

# determine the number of clusters
k_max <- 10

# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss})

# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')

## Warning: `qplot()` was deprecated in ggplot2 3.4.0.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_lifecycle_warnings()` to see where this warning was
## generated.

# k-means clustering
km <- kmeans(boston_scaled, centers = 2)

# plot the Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)

pairs(boston_scaled[1:5], col = km$cluster)

pairs(boston_scaled[6:10], col = km$cluster)

pairs(boston_scaled[11:13], col = km$cluster)

The optimal number of clusters is when the line drops a lot. Deciding on this is very subjective. One can choose two clusters for the slope being the biggest till that, or maybe 6 for there the descent evens out. We are going with 2 clusters now.

The variables separating the two groups in the pairs plots are, for instance, crim&zn and crim&nox.

Bonus section

Performing k-means on the original (standardized) Boston data:

# like above:
data("Boston")
boston_scaled <- scale(Boston)
boston_scaled <- as.data.frame(boston_scaled) # we need this too

# k-means clustering
km <- kmeans(boston_scaled, centers = 3) # trying with 3 clusters 

# plot the Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)

Performing LDA using the clusters as target classes:

# not doing a train and test split here for it was not asked for
boston_scaled$cluster <- km$cluster

# linear discriminant analysis
# lda.fit <- lda(cluster ~ ., data = boston_scaled)
# Error in lda.default(x, grouping, ...) :
# variable 4 appears to be constant within groups
# -> run the model with all variables except for the fourth one
lda.fit <- lda(cluster ~ ., data = boston_scaled[,-4])

# print the lda.fit object
lda.fit

## Call:
## lda(cluster ~ ., data = boston_scaled[, -4])
## 
## Prior probabilities of groups:
##          1          2          3 
## 0.06916996 0.61067194 0.32015810 
## 
## Group means:
##         crim         zn      indus        nox         rm        age        dis
## 1 -0.2048299 -0.1564737  0.2306535  0.3342374  0.3344149  0.3170678 -0.3634565
## 2 -0.3882449  0.2731699 -0.6264383 -0.5823006  0.2188304 -0.4585819  0.4807157
## 3  0.7847946 -0.4872402  1.1450405  1.0384727 -0.4896488  0.8062002 -0.8383961
##           rad        tax    ptratio      black      lstat       medv
## 1 -0.02700292 -0.1304164 -0.4453253  0.1787986 -0.1976385  0.6422884
## 2 -0.58641200 -0.6161585 -0.2814183  0.3151747 -0.4640135  0.3182241
## 3  1.12436056  1.2034416  0.6329916 -0.6397959  0.9277624 -0.7457491
## 
## Coefficients of linear discriminants:
##                 LD1         LD2
## crim     0.02479166 -0.13204141
## zn       0.42787622 -0.04638198
## indus    1.15646011  0.64348753
## nox      0.47943272  0.42987408
## rm       0.13637610 -0.12823804
## age     -0.06654278  0.30029385
## dis     -0.01915297  0.20848367
## rad      0.74637979  0.69845574
## tax      0.27967651 -1.02040695
## ptratio  0.19355485 -0.21964359
## black   -0.04753224  0.11581547
## lstat    0.47213016  0.02172623
## medv     0.06797263  0.92531426
## 
## Proportion of trace:
##    LD1    LD2 
## 0.9822 0.0178

Visualizing the results with a biplot:

# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  graphics::arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# target classes as numeric
classes <- as.numeric(boston_scaled$cluster)

# plot the lda results (select both lines and execute them at the same time!)
plot(lda.fit, dimen = 2)
lda.arrows(lda.fit, myscale = 1)

The variables affecting the LDs are (for example, only the top 2 listed here): indus and rad for LD1, and tax and medv for LD2. The division is not as good or clear as in some other examples encountered in the excercises. Classes 1 and 2 seem to go together, and 3 be separate, but there is overlap fbetween them all.

Super bonus section:

# scaled train data from above
data("Boston")
boston_scaled <- scale(Boston)
boston_scaled <- as.data.frame(boston_scaled) # we need this too
boston_scaled$crim <- as.numeric(boston_scaled$crim) # we need this too
crime <- cut(boston_scaled$crim, breaks = quantile(boston_scaled$crim), include.lowest = TRUE, label = c("low", "med_low", "med_high", "high"))
# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
# number of rows in the Boston dataset 
n <- nrow(boston_scaled)
# choose randomly 80% of the rows
ind <- sample(n,  size = n * 0.8)
# create train set
train <- boston_scaled[ind,]
# create test set 
test <- boston_scaled[-ind,]

# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# print the lda.fit object
# lda.fit
# has teh LD1-3 in it

# k-means clustering also needed
# km <- kmeans(train, centers = 2) # 2 clusters here to match the excercise above where we chose 2 to be the optimal number
# Warning: NAs introduced by coercionError in do_one(nmeth) : NA/NaN/Inf in foreign function call (arg 1)
# we need to make crime a numeric column (now factor)
train$crime <- as.numeric(train$crime)
km <- kmeans(train, centers = 2) # ok

# the example script copied here
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)

## [1] 404  13

dim(lda.fit$scaling)

## [1] 13  3

# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
# Next, install and access the plotly package. Create a 3D plot (cool!) of the columns of the matrix product using the code below.

# the original plot
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')

# modifying the plot: Set the color to be the crime classes of the train set
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)

# modifying the plot: color is defined by the clusters of the k-means
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = km$cluster)

The plots have some differences and similarities. In all of them, there is some overlap in the grops, and having only 2 classes in k-means still doesnt differenciate the grups that well even though in the 3D plot there are two clear groups it could pick up on. In both plots the tighter cluster has mainly one colour/group/class label, whereas the sparcer cluster has more.